Question

Acid HX is a weak acid with K_a=1.0x10^-6. A 50.0 mL sample of 1.00 M HX(aq) is titrated with 1.00 M NaOH(aq). What is the pH of the solution at the points listed below during titration?

A. pH 1.00M HX solution before titration:

B. pH after 25.0mL 1.00M NaOH added:

Please show work! Thank you!

Answer 1

A:

HX ----> H+ + X-

1M           0        0

1M-x        x        x

Ka=1 x 10^-6= x2/(1-x) -----> x= 9.99 x 10^-4 M =[H+]-----> pH= -log [H+] ----> pH= 3

B:

HX + NaOH ---->   NaX + H2O

This is a neutralization reaction, we must find the reagent that is in excess, if there is one.

molNaOH= M x V = 1M x 0.025L= 0.025 mol

molHX= 1M x 0.050L = 0.050mol

1mol HX ------- 1mol NaOH

    x         -------- 0.025 mol NaOH

x= 0.025mol HX ------> we only need 0.025 mol of HX to react with all the NaOH, so the HX is in excess. So, what is really happening is this:

HX + NaOH ----> NaX + H2O + HX excess

That HX will dissociate:

HX ----> H+ + X-

The amount of HX that is dissociating is: 0.05 mol - 0.025mol---> excess= 0.025mol HX. And the concentration will be: M= 0.025 mol/(0.050L + 0.025L)= 0.333 M

HX ----> H+ + X-

0.333      0      0

0.333-x   x      x

Ka=1 x 10^-6= x2/(0.333-x) -----> x= 5.8 x 10^-4M=[H+]----> pH=-log[H+]----> pH= 3.2